Consider an electron being ejected from the surface of a heated filament at nearly zero specd [a] through what potential difference must it be accelerated to achieve a speed of 8.4x106m/s? any relativistic effects can be neglected [b] what is the minimum b-field required to cause the electron in part [a] above to move in a circular orbit of radius r 5.0 cm?

The potential difference and the magnetic field are 200.655 V and

Explanation:

Given that,

Speed

radius = 5.0 cm

We need to calculate the potential difference

We know that,

The kinetic energy acquired by electron

Where, m = mass of  electron

q = charge

v = velocity

Put the value into the formula

(b). We need to calculate the magnetic field

Using formula of magnetic field

Put the value into the formula

Hence, The potential difference and the magnetic field are 200.655 V and

a) 200 v

b )9.555 x 10⁻⁴ T

Explanation:

Equation for electron acceleration under potential difference

q V = 1/2 mv²

V is potential diff , v is velocity after acceleration, q is charge on electron.

V= .5 x 9.1x 10⁻³¹ x( 8.4 x 10⁶)² / 1.6 x 10⁻₁¹⁹

= 200.655 V

b ) radius of circular orbit in a magnetic field of electron

r =

B )  magnetic field required =

= 9.1 x 10⁻³¹ x 8.4 x 10⁶ / 1.6 x 10⁻¹⁹ x 5 x 10⁻²

9.55 x 10⁻⁴ T

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i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!